Syntax
Insert all the implicit ( )
s and λ
s into the following abbreviated expressions. Don't just insert them freely; rather, provide the official expression, without any notational shortcuts, that is syntactically identical to the form presented.
In response to your feedback and questions, we refined the explanation of the conventions governing the use of the .
shorthand. Thanks!
x x (x x x) x
(((x x) ((x x) x)) x)
v w (\x y. v x)
((v w) (\x (\y (v x))))
(\x y. x) u v
(((\x (\y x)) u) v)
w (\x y z. x z (y z)) u v
(((w (\x (\y (\z ((x z) (y z)))))) u) v)
Mark all occurrences of (x y)
in the following terms:
(\x y. x y) x y
(\x y. x y) (x y)
\x y. x y (x y)
Reduction
Find "normal forms" for the following---that is, reduce them until no more reductions are possible. As mentioned in the notes, we'll write λx
as \x
. If we ever say "reduce" without qualifications, we mean just "beta-reduce" (as opposed to "(beta + eta)-reduce").
(\x \y. y x) z
~~>\y. y z
(\x (x x)) z
~~>z z
(\x (\x x)) z
~~>\x x
(\x (\z x)) z
~~>\y z
, be sure to change\z
to a different variable so as not to "capture"z
(\x (x (\y y))) (\z (z z))
~~>\y y
(\x (x x)) (\x (x x))
umm..., reductions will forever be possible, they just don't "do" much(\x (x x x)) (\x (x x x))
that's just mean
Booleans
For these questions, and the ones on triples below, we're setting them up so as to encourage you to experiment with Racket and to formulate your answer in Scheme/Racket syntax. But you can answer in Lambda Calculus syntax if you prefer.
Recall our definitions of true and false.
true is defined to be
\t f. t
false is defined to be\t f. f
In Racket, these functions can be defined like this:
(define true (lambda (t) (lambda (f) t)))
(define false (lambda (t) (lambda (f) f)))
(Note that they are different from Racket's primitive boolean values #t
and #f
.)
Define a
neg
operator that negatestrue
andfalse
.Expected behavior:
(((neg true) 10) 20)
evaluates to
20
, and(((neg false) 10) 20)
evaluates to
10
.(define neg (lambda (p) ((p false) true)))
Define an
or
operator.(define or (lambda (p) (lambda (q) ((p p) q))))
or:
(define or (lambda (p) (lambda (q) ((p true) q))))
Define an
xor
operator. If you haven't seen this term before, here's a truth table:true xor true == false true xor false == true false xor true == true false xor false == false
(define xor (lambda (p) (lambda (q) ((p (neg q)) q))))
Triples
Recall our definitions of ordered triples.
the triple (a, b, c) is defined to be
\f. f a b c
To extract the first element of a triple t
, you write:
t (\fst snd trd. fst)
Here are some definitions in Racket:
(define make-triple (lambda (fst) (lambda (snd) (lambda (trd) (lambda (f) (((f fst) snd) trd))))))
(define fst_of_three (lambda (fst) (lambda (snd) (lambda (trd) fst))))
(define snd_of_three (lambda (fst) (lambda (snd) (lambda (trd) snd))))
Now we can write:
(define t (((make-triple 10) 20) 30))
(t fst_of_three) ; will evaluate to 10
(t snd_of_three) ; will evaluate to 20
If you're puzzled by having the triple to the left and the function that operates on it come second, think about why it's being done this way: the triple is a package that takes a function for operating on its elements as an argument, and returns the result of operating on its elements with that function. In other words, the triple is a higher-order function.
Define the
swap12
function that permutes the elements of a triple. Expected behavior:(define t (((make-triple 10) 20) 30)) ((t swap12) fst_of_three) ; evaluates to 20 ((t swap12) snd_of_three) ; evaluates to 10
Write out the definition of
swap12
in Racket.(define swap12 (lambda (x) (lambda (y) (lambda (z) (lambda (f) (((f y) x) z))))))
Define a
dup3
function that duplicates its argument to form a triple whose elements are the same. Expected behavior:((dup3 10) fst_of_three) ; evaluates to 10 ((dup3 10) snd_of_three) ; evaluates to 10
(define dup3 (lambda (x) (lambda (f) (((f x) x) x))))
Define a
dup27
function that makes twenty-seven copies of its argument (and stores them in a data structure of your choice).OK, then we will store them in a triply-nested triple:
(define dup27 (lambda (x) (dup3 (dup3 (dup3 x)))))
Folds and Lists
Using Kapulet syntax, define
fold_left
.# fold_left (f, z) [a, b, c] == f (f (f z a) b) c letrec fold_left (f, z) xs = case xs of [] then z; x' & xs' then fold_left (f, f (z, x')) xs' end in fold_left
Using Kapulet syntax, define
filter
(problem 7 in last week's homework) in terms offold_right
and other primitive syntax likelambda
,&
, and[]
. Don't useletrec
! All theletrec
-ing that happens should come from the one inside the definition offold_right
.let filter (p, xs) = fold_right ((lambda (y, ys). if p y then y & ys else ys), []) xs in filter
Using Kapulet syntax, define
&&
in terms offold_right
. (To avoid trickiness about the infix syntax, just call itappend
.) As with problem 22 (the previous problem), don't useletrec
!let xs && ys = fold_right ((&), ys) xs # or append (xs, ys) = ... in (&&)
Using Kapulet syntax, define
head
in terms offold_right
. When applied to a non-empty list, it should give us the first element of that list. When applied to an empty list, let's say it should give us'err
. As with problem 22, don't useletrec
!let head xs = fold_right ((lambda (y, _). y), 'err) xs in head
We mentioned in the Encoding notes that
fold_left (flipped_cons, []) xs
would give us the elements ofxs
but in the reverse order. So that's how we can expressreverse
in terms offold_left
. How would you expressreverse
in terms offold_right
? As with problem 22, don't useletrec
!Here is a boring, inefficient answer
let append (ys, zs) = fold_right ((&), zs) ys; # aka (&&) f (y, prev) = append (prev, [y]); reverse xs = fold_right (f, []) xs in reverse
or (same basic idea, just written differently):
let f (y, prev) = fold_right ((&), [y]) prev; reverse xs = fold_right (f, []) xs in reverse
Here is an elegant, efficient answer following the hint
Suppose the list we want to reverse is
[10, 20, 30]
. Applyingfold_right
to this will begin by computingf (30, z)
for somef
andz
that we specify. If we made the result of that be something like30 & blah
, or any larger structure that contained something of that form, it's not clear how we could, using just the resources offold_right
, reach down into that structure and replace theblah
with some other element, as we'd evidently need to, since after the next step we should get30 & (20 & blah)
. What we'd like instead is something like this:30 & < >
Where
< >
isn't some value but rather a hole. Then with the next step, we want to plug into that hole20 & < >
, which contains its own hole. Getting:30 & (20 & < >)
And so on. That is the key to the solution. The questions you need to answer, to turn this into something executable, are:
What is a hole? How can we implement it?
A hole is a bound variable.
30 & < >
islambda x. 30 & x
.What should
f
be, so that the result of the second step, namelyf (20, 30 & < >)
, is30 & (20 & < >)
?let f (y, prev) = lambda x. prev (y & x) in ...
Given that choice of
f
, what shouldz
be, so that the result of the first step, namelyf (30, z)
is30 & < >
?The identity function:
f (30, (lambda y. y))
will reduce tolambda x. (lambda y. y) (30 & x)
, which will reduce tolambda x. 30 & x
.At the end of the
fold_right
, we're going to end with something like30 & (20 & (10 & < >))
. But what we want is[30, 20, 10]
. How can we turn what we've gotten into what we want?Supply it with
[]
as an argument.So now put it all together, and explain how to express
reverse xs
usingfold_right
and primitive syntax likelambda
,&
, and[]
?let f (y, prev) = lambda x. prev (y & x); id match lambda y. y; reverse xs = (fold_right (f, id) xs) [] in reverse
The ideas here are explored further in Chapter 8 of The Little Schemer. There they first introduce the idea of passing function as arguments to other functions, and having functions be the return values from functions. Then the
multirember&co
function discussed on pp. 137--140 (and the other...&co
functions discussed in that chapter) are more complex examples of the kind of strategy used here to definereverse
. We will be returning to these ideas and considering them more carefully later in the term.
Numbers
Given that we've agreed to Church's encoding of the numbers:
0 ≡ \f z. z
1 ≡ \f z. f z
2 ≡ \f z. f (f z)
3 ≡ \f z. f (f (f z))
...
How would you express the
succ
function in the Lambda Calculus?let succ = \n. \f z. f (n f z) in ...
Compare the definition of
cons
, which has an additional element:let cons = \d ds. \f z. f d (ds f z) in ...